It is not true. If $ n-m$ is odd $ f(-1)=0$, therefore $ f(x)=(x+1)g(x)$ with $ g(x)=x^m(x^{n-m-1}-...+1)+1$.

Rust wrote: It is not true. If $ n - m$ is odd $ f( - 1) = 0$, therefore $ f(x) = (x + 1)g(x)$ with $ g(x) = x^m(x^{n - m - 1} - ... + 1) + 1$. A/Q $ n,m$ both odd. $ n-m$ must be even.

Suppose that x^n+ x^m +x +1 is reducible x^n +x^m +x+1= P(x).Q(x) Now we 'll use an old trick Let P(x)=x.g1(x^2) +h1(x^2) , Q(x)= x.g2(x^2) + h2(x^2) Compare the odd and even degrees we must have x^2.g1(x^2).g2(x^2) + h1(x^2).h2(x^2) =1 (*) and g1(x^2).h2(x^2) + h1(x^2).g2(x^2) = x^(n-1) + x^(m-1) +1 (**) From (*) we get t.g1(t).g2(t) + h1(t).h2(t)=1 From (**) we get g1(t).h2(t) +h1(t).g2(t)=t^p +t^q +1 ( where p=n-1/2, q= m-1/2 ) Hence t.[t^p+t^q+1]^2 - 1^2 = - [ t.g1(t)^2 - h1(t)^2 ] * [ t.g2(t)^2 - h1(t)^2 ] .....................

I'm blocked here, anyone could continue in this way or propose another solution?

See Ljunggren W, On the irreducibility of certain trinomials and quadrinomials, Math.Scand. 8 (1960) Main idea is if $ f$ is not irreducible, then there exist its divisor $ g(x)$ such that $ g(x)=x^{deg(g)}\cdot g(x^{-1})$. This leads $ f(x)$ has a root $ a$ such that $ a^{-1}$ is its another root. After some calculation, $ a$ must be even and odd degree root of $ -1$, that leads to contradiction.

Proof of the mentioned result is quite long and boring. I doubt it can be invented during TST. I would like to see something simple in case $ m$ and $ n$ are odd.

any solution, please

any one ??

http://www.mscand.dk/article/download/10593/8614

Here's an explicit write up for this case. Suppose that FTSOC $f(x) = g(x)h(x)$. Define $\tilde{f}$ as the reciprocal polynomial. Then if $k = g \tilde{h}$, $k \ne \pm f, \pm \tilde{f}$ and \[ k\tilde{k} = f\tilde{f} = (x^n + x^m + x + 1)(x^n + x^{n-1} + x^{n-m} + 1) \]By considering the coefficient of $x^n$, the sum of square terms in $k$ is the same as $f$ and thus $4$. WLOG suppose that $k$ is monic. Let $k = x^n \pm x^{u} \pm x^{v} \pm 1$ for $u, v < n$. Then, all coefficients must be positive since $f\tilde{f}$ has $16$ positive coefficients. As such, \[ (x^n + x^m + x + 1)(x^n + x^{n-1} + x^{n-m} + 1) = (x^n + x^{u} + x^{v} + 1)(x^n + x^{n-v} + x^{n-u} + 1) \]It follows that $u + v = m + 1$ must hold by considering the sum of exponents. Now, consider the $x^{2n - 1}$ term which appears in the LHS. This only holds if one of $u, v$ is either $n - 1$ or $1$. This implies that $k$ is either $f$ or $\tilde{f}$, contradiction.